Download Chapter 8 Getting Things to Move: Momentum and Kinetic Energy

Chapter 8
Getting Things to Move:
Momentum and Kinetic Energy
In This Chapter
Working with impulse
Gaining momentum
Conserving momentum
Conserving kinetic energy
Handling two-dimensional collisions
T
hings collide all the time, and when they do, physics gets involved. Knowing about the
conservation of momentum and kinetic energy lets you handle this kind of problem —
when things hit each other, you can predict what’s going to happen.
The conservation of momentum and kinetic energy uses two of the strongest tools that
physics has in the physical world, and this chapter helps you become a pro at working with
these tools.
Acting on Impulse
When you apply a force for a certain amount of time, you create an impulse. In fact, that’s
the definition of impulse — impulse equals the force applied multiplied by the time it was
applied. Here’s the equation:
Impulse = F t
Note that this is a vector equation (for a review of vector equations, check out Chapter 3).
Impulse can be an important quantity when you’re solving physics problems because you
can relate impulse to momentum (which I cover in the next section), and working with
momentum is how you solve collision problems in physics.
Here’s an example of impulse in action: You’re playing pool, and you strike a pool ball with
the cue. The cue may be in contact with the ball for only a millisecond, but there’s an observable result — the ball flies off in the opposite direction. That result is the impulse.
What are the units of impulse? You have force multiplied by time, so the unit is the
Newton-second.
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Q.
Suppose that you’re playing pool and hit
a pool ball for 10.0 milliseconds (a millisecond is 1⁄1000 of a second) with a force of
20.0 N. What impulse did you impart to the
pool ball?
A.
The correct answer is 0.2 N-sec, in the
direction of the force.
1. Use the equation Impulse = F t.
2. Plug in the numbers:
Impulse = F t = (20)·(1.0 × 10–2) = 0.20 N-sec,
in the direction of the force.
1.
You’re disgusted with your computer and
give it a whack. If your hand is in contact
with the computer for 100.0 milliseconds
with a force of 100.0 N, what impulse do
you impart to the computer?
Solve It
2.
You’re standing under the eaves of your
house when a huge icicle breaks off and
hits you, imparting a force of 300.0 N for
0.10 seconds. What was the impulse?
Solve It
Chapter 8: Getting Things to Move: Momentum and Kinetic Energy
Getting Some Momentum
Momentum is the most important quantity when it comes to handling collisions in
physics. Momentum is a physical quantity defined as the product of mass multiplied
by velocity. Note that that’s velocity, not speed, so momentum is a vector quantity. Its
symbol is p; here’s the equation for momentum:
p=mv
Note that p is always in the same direction as v because m is a scalar value (that is, a
single value, not a value with multiple components like a vector). It turns out that
momentum is conserved in collisions, which means that the momentum before a collision is the same as the momentum after a collision (assuming that minimal heat was generated by the collision — that little energy went into deforming the colliding objects). So
if you know the original momentum in the collision, you can predict what the situation
will be after the collision (and physicists are always delighted by such predictions).
What are the units of momentum? Momentum is mass times velocity, so its unit is
kilogram·meters/second (kg-m/sec) in the MKS system.
Q.
Suppose that you’re in an 800.0 kg race car
going 200.0 miles an hour due east. If you
have a mass of 60.0 kg, what is the total
momentum?
A.
The correct answer is 2.5 × 105 kg-m/sec,
due east.
1. Use the equation p = m v.
2. Plug in the numbers, assuming that
200 miles per hour is about 89.4 m/sec:
p = m v = (800 + 60)·(89.4) = 1.43 × 105
kg-m/sec, due east.
3.
You’re running north at 3.0 m/sec. If you
have a mass of 80.0 kg, what is your
momentum?
Solve It
4.
You’re falling out of an airplane, and before
opening your parachute, you hit a speed of
100.0 m/sec. What is your momentum if
you have a mass of 80.0 kg?
Solve It
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5.
You’re pushing a 10.0 kg box of dishes
across the kitchen floor at a rate of 4.0
m/sec. What is its momentum?
Solve It
6.
You’re pushing an 800.0 kg car down the
road, and it’s going at 6.0 m/sec west. How
much momentum does it have?
Solve It
Relating Impulse and Momentum
It turns out that there’s a direct connection between impulse and momentum. If you hit
a pool ball with a cue, the cue imparts a certain impulse to the ball, causing the ball to
end up with a particular momentum.
How can you relate impulse to momentum? Easy. The impulse you impart to an object
gives it a change in momentum equal to that impulse, so:
Impulse = F·t = p = m·v
Q.
A.
If you hit a stationary 400.0 g pool ball with
a force of 100.0 N for 0.10 seconds, what is
its final speed?
The correct answer is 25 m/sec.
1. Use the equation F·t = p.
2. Find the impulse first:
Impulse = F·t = (100)·(0.1) = 10 N-sec
3. That impulse becomes the pool ball’s
new momentum:
p = (100)·(0.1) = 10 kg-m/sec
4. Momentum equals mass times velocity,
so to find speed (the magnitude of velocity), you can solve for v, like so:
v = p / m = (10) / 0.4) = 25 m/sec
Chapter 8: Getting Things to Move: Momentum and Kinetic Energy
7.
You hit a hockey puck, mass 450 g, with a
force of 150 N for 0.10 seconds. If it started
at rest, what is its final speed?
Solve It
9.
You kick a 1.0 kg soccer ball with a force of
400.0 N for 0.20 seconds. What is its final
speed?
Solve It
8.
You’re standing on an ice rink when
another skater hits you, imparting a force
of 200.0 N for 0.20 seconds. If you have a
mass of 90.0 kg, what is your final speed?
Solve It
10.
You hit a 400 g baseball with a force of
400.0 N for 0.10 seconds. The baseball was
traveling toward you at 40 m/sec. What is
its final speed?
Solve It
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Part III: Being Energetic: Work
Conserving Momentum
The major tool you have in calculating what’s going to happen in collisions is the
knowledge that momentum is conserved. You know that the total momentum before
the collision is the same as the total momentum after the collision as long as there are
no significant outside forces.
When you have two objects that collide (one is initially at rest and the other is moving),
and you know the final velocity and mass of one object after the collision, you can calculate the final velocity of the other object. You can do this because the total momentum is
conserved, so it’s the same, before and after the collision, as shown by this equation:
po = pf
Q.
A.
A pool ball with a mass of 400 g and a
speed of 30 m/sec hits another pool ball
that’s at rest. If the first pool ball ends up
going in the same direction with a speed
of 10 m/sec, what is the new speed of the
second pool ball?
3. The original total momentum equals the
total final momentum, which is given by
this expression, where p2f is the final
momentum of the second ball:
The correct answer is 20 m/sec.
4. Solve for p2f:
po = 12 kg-m/sec = pf = (0.4)·(10) + p2f
p2f = 12 – 4.0 = 8.0 kg-m/sec
1. Use the equation po = pf.
5. Divide p2f by the pool ball’s mass to find
its speed:
2. All travel is in the same direction in
this question, so you can treat it as a
scalar equation. Find the original total
momentum:
p2f / m = 8.0 / 0.4 = 20. m/sec
po = (0.4)·(30) = 12 kg-m/sec
11.
A 450 g hockey puck traveling at 60 m/sec
hits a stationary puck. If the first puck ends
up going at 20 m/sec in the same direction
as the second puck, how fast is the second
puck moving?
Solve It
12.
You’re driving a bumper car at a circus at
18 m/sec, and you hit another car that’s at
rest. If you end up going at 6.0 m/sec, what
is the final speed of the other car, given
that both cars have 100.0 kg mass, you
have 80.0 kg mass, and the other person
has a mass of 70.0 kg?
Solve It
Chapter 8: Getting Things to Move: Momentum and Kinetic Energy
13.
On the athletic field, a golf ball with a mass
of 0.20 kg and speed of 100.0 m/sec hits a
soccer ball at rest that has a mass of 1.0 kg.
If the golf ball ends up at rest, what is the
soccer ball’s final speed, given that it travels in the same direction as the golf ball
was originally traveling?
Solve It
14.
You’re stopped at a traffic light when a
1000.0 kg car (including driver) hits you
from behind at 40.0 m/sec. Ouch. If the
other car ends up moving at 12 m/sec, and
if you and your car have a mass of 940 kg,
what is your final speed?
Solve It
Conserving Kinetic Energy — or Not
In some types of collisions, called elastic collisions, kinetic energy and momentum are
conserved. Here are the equations for the conservation of these factors:
KEo = KEf
po = pf
Check out this idea in action: Suppose that you’re in a car when you hit the car in front
of you (elastically — no deformation of bumpers is involved), which started at rest.
You know that momentum is always conserved, and you know that the car in front of
you was stopped when you hit it, so if your car is car 1 and the other one is car 2, you
get this equation:
m1vf1 +m2vf2 = m1vo1
This equation can’t tell you what vf1 and vf2 are, because there are two unknowns and
only one equation. You can’t solve for either vf1 or vf2 exactly in this case, even if you
know the masses and vo1. So to solve for both final speeds, you need another equation
to constrain what’s going on here. That means using the conservation of kinetic
energy. The collision was an elastic one, so kinetic energy was indeed conserved. That
means that:
1 m v 2+ 1 m v 2= 1 m v 2
2 1 f1 2 2 f2 2 1 o1
With two equations and two unknowns, vf1 and vf2, you can solve for those unknowns
in terms of the masses and vo1.
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You probably won’t be asked to solve questions of this kind on physics tests because, in
addition to being two simultaneous equations, the second equation has a lot of squared
velocities in it. But it’s one you may see in homework. When you do the math, you get
v f1 =
_ m 1 - m 2 i $ v o1
_m1 + m 2 i
and
v f2 =
2 $ m 1 $ v o1
_m1 + m 2 i
This is a more substantial result than you get from problems that just use the conservation of momentum; in such problems, you can solve for only one final speed. Here,
using both the conservation of momentum and kinetic energy, you can solve for both
objects’ final speeds.
Q.
A.
You’re in a car that hits the at-rest car in
front of you. If you and your car’s mass is
1000.0 kg, the mass of the car and driver
ahead of you is 900.0 kg, and if you started
at 44 m/sec, what are the final speeds of
the two cars? Assume that all the action
happens in the same line as your original
direction of travel.
v f1 =
v f1 =
v f2 =
2 $ m 1 $ v o1
_m1 + m 2 i
4. Plug in the numbers:
v f2 =
2 $ m 1 $ v o1 2 $ ^1000h $ ^ 44h
=
= 46 m/ sec
^1900h
_m1 + m 2 i
_ m 1 - m 2 i $ v o1
_m1 + m 2 i
A 450 g hockey puck traveling at 60.0 m/sec
hits a stationary puck with the same mass.
What are the final speeds of the pucks,
given that the collision is elastic and that
all motion takes place along the same line?
Solve It
_ m 1 - m 2 i $ v o1 ^100h $ ^ 44h
=
= 2.3 m/ sec
^1900h
_m1 + m 2 i
3. Use this equation to find the final speed
of the other car:
The correct answer is that your car
moves 2.3 m/sec, and the other car moves
46 m/sec.
1. Use this equation to find the final speed
of your car:
15.
2. Plug in the numbers:
16.
You’re driving a bumper car at 23 m/sec,
and you hit another bumper car that’s at
rest. If you and your car have a mass of
300 kg, and the mass of the other car and
driver is 240 kg, what are the final speeds
of the cars?
Solve It
Chapter 8: Getting Things to Move: Momentum and Kinetic Energy
Collisions in Two Dimensions
Collisions can take place in two dimensions. For example, soccer balls can move any
which way on a soccer field, not just along a single line. Soccer balls can end up going
north or south, east or west, or a combination of those — not just along the east-west
axis. So you have to be prepared to handle collisions in two dimensions.
Q.
In Figure 8-1, there’s been an accident at an
Italian restaurant, and two meatballs are
colliding. Assuming vo1 = 10.0 m/sec,
vo2 = 5.0 m/sec, vf2 = 6.0 m/sec, and the
masses of the meatballs are equal, what
are θ and vf1?
vf1
θ
1
1
40°
vo1
1
2
vo2
2
Figure 8-1:
Colliding
objects.
A.
2
30°
vf2
The correct answer is θ =24° and vf1 =
8.2 m/sec.
1. You can assume that collisions between
meatballs don’t conserve kinetic energy.
However, momentum is conserved. In
fact, momentum is conserved in both the
x and y directions, which means that:
pfx = pox
and:
pfy = poy
2. Here’s what the original momentum in
the x direction was:
pfx = pox = m1·vo1·cos 40° + m2 vo1
3. Momentum is conserved in the x direction, so:
pfx = pox = m1·vo1·cos 40° + m2 vo2 =
m1·vf1x + m2 vf2 cos 30°
4. Which means that:
m1·vf1x = m1·vo1·cos 40° +
m2 vo2 – m2 vf2 cos 30°
5. Divide by m1:
v f1x =
m 1 $ v o1 $ cos 40c + m 2 v o2 - m 2 v f2 cos 30c
m1
And because m1 = m2, this becomes
vf1x = vo1·cos 40° + vo2 – vf2 cos 30°
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Part III: Being Energetic: Work
6. Plug in the numbers:
vf1x = vo1·cos 40° + vo2 – vf2 cos 30° =
(10) (.766) + 5.0 – (6.0) (.866) = 7.5 m/sec
7. Now for the y direction. Here’s what the
original momentum in the y direction
looks like:
10. Because the two masses are equal, this
becomes
vf1y = vo1·sin 40° – vf2 sin 30°
11. Plug in the numbers:
vf1y = vo1·sin 40° – vf2 sin 30° =
(10) (.642) – (6.0) (0.5) = 3.4 m/sec
pfy = poy = m1·vo1·sin 40°
12. So:
8. Set that equal to the final momentum in
the y direction:
pfx = pox = m1·vo1·sin 40° =
m1·vf1y + m2 vf2 sin 30°
That equation turns into:
m1vf1y = m1·vo1·sin 40° – m2 vf2 sin 30°
9. Solve for the final velocity component of
meatball 1’s y velocity:
v f1y =
17.
vf1x = 7.5 m/sec
vf1y = 3.4 m/sec
That means that the angle θ is
θ = tan–1(3.4/7.5) = tan–1(0.45) = 24°
And the magnitude of vf1 is
2
2
v f1 = v f1x + v f1y = 7.5 2 + 3.4 2 = 8.2 m/s
m 1 $ v o1 $ sin 40c - m 2 v f2 sin 30c
m1
Assume that the two objects in Figure 8-1 are hockey pucks of equal mass. Assuming that
vo1 = 15 m/sec, vo2 = 7.0 m/sec, and vf2 = 7.0 m/sec, what are θ and vf1, assuming that momentum
is conserved but kinetic energy is not?
Solve It
Chapter 8: Getting Things to Move: Momentum and Kinetic Energy
18.
Assume that the two objects in Figure 8-2 are tennis balls of equal mass. Assuming that
vo1 = 12 m/sec, vo2 = 8.0 m/sec, and vf2 = 6.0 m/sec, what are θ and vf1, assuming that momentum
is conserved but kinetic energy is not?
Solve It
vf1
θ
1
1
35°
vo1
1
2
Figure 8-2:
Dragging a
mass.
vo2
2
2
42°
vf2
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